Selected quad for the lemma: end_n

Word A	Word B	Word C	Word D	Occurrence	Frequency Band	MI	MI Band	Prominent
end_n	foot_n	half_a	inch_n	1,859	5	11.0254	5	true

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ID	Title	Author	Corrected Date of Publication (TCP Date of Publication)	STC	Words	Pages
A78143	The measurer's guide: or, the whole art of measuring made short, plain and easie Shewing, 1. How to measure any plain superficies. 2. How to measure all sorts of regular solids. 3. The art of gaugeing. 4. How to measure artificers work, viz. carpenters, joyners, plasterers, painters, paviers, glaziers, bricklayers, tylors, &c. of singuler use to all gentlmen, artificers and others. By John Barker.	Barker, John, writer on fortification.	1692 (1692)	Wing B771A; ESTC R228729	29,851	150

to Measure Tyling pag. 128 ERRATA Pag. 1. in the Title Instead of Superfices Read Superficies P. 54. L. 5. for 44. 64 Inches Read 4464. P. 25. L 5. for Superfices Read Superficies THE MEASURER's GUIDE OR THE Whole ART OF Measuring MADE Short Plain and Easie CHAP. I. How to measure any Plain Superfices PROP. I. HOW to find the Superficial Content of a Geometrical Square Example Admit there is a Square represented by the Figure A B C D whose sides are each Nine Feet whose Area or Superficial Content is required To find which multiply its side 9 by its self and the product will be 81 Feet which is the Content of that Square PROP. II. How to find the Superficial Content of a Parallelogram or long Square Multiply the length by the breadth and the Product will be the Content thereof Example Admit there is a long Square represented by the Figure B C D E whose length B C is Twenty Two Inches and breadth B D Twelve Inches whose Content is required To find which multiply the length 22 by the breadth 12 and the Product will be 264 Inches which is the Content of the long Square required which you may reduce into Feet by dividing the same by 144 the Number of Inches contained in a superficial Foot and the Quotient will be 1 Foot and 120 will remain with 120 144 of a Foot which is 10 Inches But this way of Working being very tedious and troublesome by reason of the Reduction for your Multiplier and Multiplicand must be reduced both into the Lowest Denomination and afterwards reduced into Feet I would advise the young Artist to work by Decimals and for that end to provide himself of a Two-Foot Rule on which each Foot shall be divided into Ten equal parts and each of those parts into Ten other equal parts So is the Foot divided into 100 equal parts and thereby is fit to take the Dimensions in Feet and parts of a Foot by which Rule if the Dimensions of the Square in the last Example was taken the length B C would be 1. 83 Feet and the breadth B D 1 Foot Then if you multiply the length by the breadth as before the Product will give 1. 83 or 1 Foot 10 Inches very near So is the Content found without any more trouble which Method I have made use of in some of the Examples in the following Discourse of Measuring of Superfices and Solids but in such sort as any person that understands not Decimal Arithmetick may easily apply them to Vulgar Arithmetick But generally I have given the Examples in Integers or Feet If it happens that any long Square Superfices be broader at one end than at the other whose Content is required first add the two ends together and take half their Sum for the true breadth and multiply it into the length and the Productwill be the Content thereof Example Admit there is a peice of Glass or any other Superfices that is One Foot broad at one end and Two Feet broad at the other and Six Feet long whose Content is required To perform which add the two ends together viz. One Foot and Two Feet and their Sum will be Three Feet the half of which is 1. 5 Foot or One Foot Six Inches which multiply into the length Six Feet and the Product will give Nine Feet for the Content required PROP. III. How to find how much in Length of any Long Square Superfices goes to make a Superficial Foot Divide 1 by the given breadth and the Quotient will give how much of the length of any Superfices goes to make a Superficial Foot thereof Example Admit there is a long square piece of Board or any other Superfices whose breadth is 1. 32 feet and it be required to know how far one ought to measure along the same to make a superficial foot thereof To perform which divide 1 by the breadth 132 and the Quotient will give 757 which is Nine Inches and something more and so much of the Length goes to make a superficial foot as in the following Work PROP. IV. How to find the superficial Content of a Right lined Triangle Although Right lined Triangles are of several Kinds and Forms as first in respect of their Angles they are either Right Angled or Oblique Angled Acute Angled or Obtuse Angled Second in respect of their sides They are either an Equilateral Isosceles or Scalenium Triangle But seeing they are all measured by one and the same Rule I shall add but One Example for all and take this for a general Rule Multiply half the Perpendicular into the Base and the Product will be the Content of the Triangle or Multiply half the Base into the whole Perpendicular and the Product will give the same thing Or if you Multiply the whole Base into the whole Perpendicular half the Product is the Content of the Triangle Example Admit there is a Triangle A B C whose Base A C is 72 feet and the Perpendicular B Q is 36 feet whose Content is required To find which multiply half the Perpendicular which is 18 feet into the Base 72 feet and the Product will be 1296 feet which is the Content of the said Triangle Or thus Multiply half the Base which is 36 feet into the Perpendicular 36 feet and the Product will give 1296 feet as before for the Content of the said Triangle PROP. V. How to find the Superficial Content of a Polygon or many equal sided Superfices Multiply the Length of the Perpendicular let fall from the Center to the middle of one of the sides into half the Sum of the sides or half the Perpendicular into the Sum of the Sides and the Product shall be the Content of Polygon Example Admit the Polygon given be a Hexagon represented by the Figure A B C D E F or a Six equal sided Figure whose side is 19 and the Perpendicular or ☉ G is 16 feet Multiply half the Sum of the sides which is 57 by the length of the Perpendicular 16 and the Product will be 912 feet which is the Content of the Polygon required Or Multiply the Sum of all the sides which is 114 by half the Perpendicular 8 and the Product will give 912 as before for the Content of the said Polygon The reason of this manner of Working is very plain if from the Center you draw the Lines ☉ E and ☉ D thereby making the Triangle E ☉ D whose Content by the third Proposition is found by Multiplying the Perpendicular ☉ G 16 into half the side E D which is 9.5 and there being in the Hexagon 6 such Triangles therefore the Perpendicular ☉ G being Multiplied into 6 times half the side D E produceth 912 for the Content as before PROP. VI. How to find the Superficial Content of a Trapezium or Table A Trapezium is a Figure consisting of four unequal sides and as may unequal Angles as is the Figure following A B C D. To Measure which

and breadth B D 40 whose superficial content is required To find which Multiply the length 50 by the breadth 40 and the Product will be 2000 which Multiply by 11 and the Product will be 22000 which Divide by 14 and the Quotient will be 1571 which is the content of the Oval required CHAP. II. How to Measure all sorts of Regular Solids HAving in the last Chapter shewed you how to Measure any Plain Superfices which if well understood the content of all Solid Bodies whose ends are of the same Dimensions will be found by the following Rule First find the superficial content at the end by some of the Propositions foregoing and Multiply the same by the length and the Product will give the solid content PROP. I. How to find the solid Content of a Cube Multiply its side into its self and that Product Multiply by the side again and that Product will be the solid content of the cube desired Example Admit there is a Cube whose side is 6 Feet represented by the Figure A B C D E F G whose solid content is required To find which Multiply the side 6 by its self and the Product will be 36 which again Multiply by 6 and the Product will be 216 Feet which is the solid content of the said Cube PROP. II. How to find the Solid Content of a Parallelepipedon or long Square First find the superficial content at the end by Propositions the first or second of the last Chapter which Multiply into the length and the Product will be the solid content Example Admit there is a Parallelepipedon represented by the Figure H I K L M N O whose Square at the end is 3 feet long and 2 feet broad and its length K M is 39 feet whose solid content is required To find which Multiply the length at the top H I 3 by the breadth H K 2 and the Product will give 6 which is the superficial content at the top which Multiply into the length K M 39 feet and the Product will be 234 feet which is the solid content of the said Parallelepipedon PROP. III. How to find the solid Content of a Cylinder First find the superficial content of the circle at the top or base by Proposition the 10 11 or 12 of the last Chapter then Multiply it by the length and the Product will be the solid content thereof Example Admit there is a Cylinder represented by the Figure G whose circumference of the circle at the end or base is 44 Inches and the length of the Cylinder 16 Inches whose solid content is required To find which first find the superficial Content of the Circle at the Top or Base by Proposition the 11th of the last Chapter which you will find to be 154 Inches which multiply by the length 16 Inches and the Product will be 2464 Inches which is the Content of the Cylinder required which you may reduce into feet by dividing the same by 1728 the number of Inches contained in a solid Foot and the Quotient will give 1 Foot and there will remain 636 which is 736 1728 of a Foot But as I said before the best way is to work by Decimals but I have added this Example for their sakes that understand not Decimals But this as well as any other Question may be better resolved by Cross Multiplication than by this Common Method to perform which if the Learner understands it not already I refer him to the last Chapter of this Book and you will find the Answer to be 1 Foot 4 Inches and 10 Seconds as in the following Work But here I presume it may not be improper to shew the great Error that common Measurers are guilty of in the measuring of Round Timber whose Method is thus they girt the Tree or peice of Timber about and take the one fourth part of the Circumference for the true Square which is very erroneous as may appear by the last Example where the Circumference or Girt of the Cylinder was 44 and the Content of the Circle 154. Whereas if you had taken the one fourth of the Circumference 44 which is 11 for the Square Root of the Circle which multiplied into its self the Product is but 121 whereas the Content of the Circle is found to be 154 from which substract 121 and the Remainder will be 33. So much doth the common-way want of the true Content of the Circle from whence it is evident that according to the common way of measuring there is lost something above the ● 5 part of any Round Timber so measured But all that can be said in Defence of this Custom is That all Trees growing round must be hewed square before it can be fit for any use so this Advantage in measuring may be very well allowed for what goes to waste in chips they being good for nothing but the fire and although Measurers think not of this Excuse but take their Rule for an absolute Truth I presume was the first Occasion of the Rule though it is a very Unreasonable Allowance as may thus appear the inscribed Square of the circle of the Cylinder in the last Example will be 10 Inches as you may prove by the following Proportion As 1.000 is to 0.225 so is the circumference 44 to the inscribed Square 9.900 which wants but very little of 10 inches and this is the greatest Square such a Round piece of Timber can be hew'n into and this multiplied into its self gives 98 inches and .010 parts for its content Now if you add the content of the inscribed Square 98 and the true content of the circle 154 together their Sum will be 252 and the Middle or Mean thereof which is half their Sum is 126 and the content after the common way was but 121. So that you may see this Rule gives an indifferent Allowance both to the Buyer and Seller But I presume notwithstanding all that hath been said that the true content ought to be given and the Measure exactly known And as to the Waste thereof Goodness or Badness of the Timber there ought to be an allowance made in the Price PROP. IV. How to know how much in length of any Solid Body having equal Bases goes to make a Solid Foot thereof First find the superficial content at the End or Base by which divide 1 to wit 1 solid foot and the Quotient will be the Length that goes to make a foot solid thereof Example Admit there is a piece of Timber or any other solid body that is terminated at each end by two equal Geometrical Squares whose side is 1. 55 and it be required to know how much of the Length thereof goes to make a foot solid To perform which first find the superficial content at the End or Base by Proposition the 1st of the last Chapter which you will find to be 2. 4025 by which divide 1 and the Quotient will give .416 foot or 5 inches as in the following Work

in the former Example and you will find the Content of the same to be 10 Squarse 7 Inches as you see in the Following Work How to Measure the Roof of any Building Example Admit there is a Building that is 93 Feet Long from the out-side of the Wall at one end to the out-side of the Wall at the other end and the Breadth from the out-side of one Wall to the out-side of the other Wall 36 Feet and it be required to know how many Squares of Roofing there is in that Building To perform which if the Building be true Roof'd you need not Measure the Length of the Rafters or the side-pieces but take this for a General Rule viz. that the Length of each Rafter is three Fourths of the Breadth of any Building so in this Example the Breadth of the Building being 36 Feet ¾ thereof is 27 Feet which is the Length of each Rafter or Breadth of one side of the Roof which being doubled for so it must be for both the sides is 54 Feet which you must Multiply by the Length of the Building 93 and the Product will be 5022 feet for the Content of the said Roof which reduce into Squares by dividing the same by 100 so will you find 50 Squares and 2 feet and 20 100 of a foot to be the Content of the Roof required as in the following work How to Measure the sides of a Timber Building Example Admit there is a Timber Building that is 36 feet broad at each end and 93 feet long and the height to the Rafters or side of the Roof is 69 feet whose Content is required To perform which because the two sides are equal and likewise the two ends are equal you need not stand to find the Content of each side severally but add all the four sides together whose sum will be 258 feet which Multiply by the height of the Building 69 and the product will be 17802 which reduce into squares so will you find the Content of all the four sides of the said wod Building to be 178 squares and 2 100 of a square as in the following Work How to Measure the Gable-end of a Timber Building Example Admit it be required to find the Content of the Gable-end of a Building whose breadth is 72 feet from the out-side of one Wall to the out-side of the other the gable-Gable-end of which Building is represented by the following Triangle ABC whose Base A C is 72 feet equal to the Breadth of the Building and the Perpendicular B Q is 36 feet To find the Content of which Multiply 72 the Base by 18 halve the Perpendicular and the Product will give 1296 feet which reduce into squares so will you find the Content of the said gable-Gable-end to be 12 squares and 9 feet and 6 Inches How to Measure Paving Painting Wainscoting and Plastering c. These are all Measured by the Yard Square containing 9 feet so that in Measuring of this sort of Work all that it differs from the former is only instead of Reducing the Content into Squares by dividing the same by 100 you must Reduce it into Square Yards by dividing the feet by 9 the number of feet in a square Yard Example 1. Admit there is a Court that is Paved containing 76 feet 6 Inches in Length and 17 feet and four Inches in Breadth whose Content is required To find which Multiply the length of the said Court 76 feet and 6 inches by the breadth 17 feet and 4 Inches as before Taught and the Product will be 136 feet which feet Divide by 9 the number of feet in a Square Yard and the Quotient will be 147 and 3 will Remain which is of a ● Yard so is the Content of the said Court found to be 147 Square Yards and 3 9 of a Yard or one foot as in the following Work Example 2. Admit there is a piece of Wainscot that is 6 foot long and 3 feet 6 inches Broad whose Content is required To perform which work in every respect as in the foregoing Example and the Answer will be 2 square Yards and ● of a Yard or 1 foot as in the following Work Example 3. Admit there is a Room that is 9 feet 6 inches long and 8 feet Broad and 7 feet high which is Painted whose Content is required To perform which first Multiply the breadth by the height and the Product will be the Content of one end of the Room which because the two ends are alike double the same and it will give the Content of both the ends then also Multiply the length of the Room by the height and the Product will be the Content of one of the sides which being doubled because the sides are both alike gives the Content of both sides which being added to the Content of both the ends before found the sum is the Content of the 4 sides of the Room required which reduce into square yards as before taught by Dividing the same by 9 and the Answer will be 27 square yards and 2 9 of a yard * ⁎ * * ⁎ * Note that the Demensions of a Room that is Painted is best taken by a String because of the Mouldings which you bend as you please and so Measure the Mouldings with the rest which must be done as being part of the Painting Example Admit there is a Room that is Plaistered that is 16 Feet 8 Inches Long and 10 Feet 6 Inches Broad and 7 Feet 11 Inches High whose Content is Required To perform which find the Contents of the 4 sides of the Rooms as directed in the last Example which you will find to be 430 Feet 1 Inch and 8 Seconds then find the Content of the Top of the Room which is done by Multiplying the Length of the Room by the Breadth viz. 16 Feet and 8 Inches by 10 Feet and 6 Inches and the Product will be 175 Feet which Add to the Content of the sides and their Sum is the Content of the said Room which you will find to be 605 Feet 1 Inch and 8 Seconds which Reduce into Yards as before taught and so will the Answer be 67 Square Yards and 2 9 of a Yard As in the following Work Note that in Measuring of Plaisterers Work there is no allowance made for the Windows Doors or Chimnies But here I judge it may not be improper to add how to find the Superficial Content of Solid Bodies because Painters do very frequently Paint such Bodies How to find the Superficial Content of any Solid Body First How to find the Superficial Content of a Cylinder Multiply the Length by the Circumference and the Product will be the Superficial Content desired Example Admit there is a Round Pillar that is Painted whose Circumference is 5 Feet 10 Inches and the Length thereof is 12 Feet 8 Inches and it be required to know how many Yards Square of Painting there is in the same To perform which Multiply the

PROP. V. How to find the solid Content of a Cone First find the superficial content of the circle at the Base by Proposition the 10th 11th or 12th of the last Chapter which multiply by ½ of the heighth and the Product will be the solid content of the Cone Example Admit there is a Cone represented by the Figure B whose Diametre at the Base is 5 feet and its Heighth or Altitude is 18 feet whose solid Content is required To find which first find the superficial content of the Base by Proposition the 11th of the last Chapter which you will find to be 19. 64 which multiply by ⅓ of the heighth which is 6 and the Product will give 117. 84 feet which is the solid content of the Cube required PROP. VI. How to find the Solid Content of a Pyramid A Pyramid is a Silid comprehended under plain Surfaces and forms a Triangular Quadrangular or any Mutangular Base diminishing equally less and less till it diminishes in a Point at the Top as a Cone To find the solid content of any such Figure first find the superficial content of the Base by some of the Propositions of the last Chapter which multiply into ⅓ of its Altitude or Heighth and the Product will be the solid content thereof Example Admit there is a Pyramid represented by the Figure M whose Base is a Hexagon the side of which is 25 feet and its Perpendicular heighth is 60 feet whose solid content is required To perform which first find the superficial content of the Base by Proposition 5. of the last Chapter viz. by multiplying half the Sum of the sides which is 75 feet by the Perpendicular let fall from the Center to the midst of one of the sides which will be 13 feet and the Product will give 975 feet for the superficial content of the Base which multiply by ⅓ of the heighth 20 and the Product 19500 feet for the solid content of the Pyramid required PROP. VII How to find the Solid Content of the Segment of a Cone or Pyramid Here note that all tapering Timber or Stone c. whose Bases are Regular Figures are Segments of either a Cone or Pyramid To find the solid content of any such Segment first find the superficial content of both the Bases by some of the Propositions of the last Chapter then multiply the superficial content of the greater Base by the superficial content of the lesser and extract the square Root of that Product then add that square Root and the superficial content of the Two Bases together and their Sum multiply by ⅓ part of the Length and that Product will give the solid content of the Segment required Example Admit there is a Segment of a Pyramid whose Bases are Squares represented by the Figure P G and the side of the great Square at G is 1.5 foot or 1 foot 6 Inches and the side of the lesser at P is .5 foot or 6 Inches and the Length of the said Segment is 30 feet whose solid content is required To find which first find the superficial content of both the Ends by Proposition the first of the last Chapter the greater of which you will find to be 2.25 feet and the lesser 0.25 foot then multiply the superficial content at the greater end at G 2.25 by the superficial content of the lesser at P .25 and the Product will be .5625 feet then extract the square Root of that Product which you will find to be .75 feet to which add the superficial content of the Two Ends and their Sum will be 3.25 which multiply by ⅓ part of the Length which is 10 feet and the Product will be 32.50 feet which is the solid content of the said Segment required as in the following work If you work the same by Vulgar Arithmetick you will find the content to be 32 feet 6 Inches PROP. VIII How to find the Solid Content of a Globe or Sphere the Diametre being given To find which work by the following Proportion As 6 times 7 which is 42 Is to 22 So is the Cube of the Diametre of any Globe To the solid content thereof Example Admit there is a Globe or Sphere represented by the Figure following whose Diametre is 24 Inches and be required to find the solid content thereof To perform which first cube the Diametre 24 Inches whose Cube you will find to be 1.3824 Inches then work by the before-going Proportion and say As 42 is to 22 so is 13824 to the solid content of the said Globe which you will find to be 7241 6 42 Inches PROP. IX How to find the Solid Content of a Globe or Sphere the Circumference being given To find which work by the following Proportion As 1.0000 Is to 0.16887 So is the Cube of the Circumference of any Globe To the solid content thereof PROP. X. How to measure any Body that is hollow Admit it were required to find the solid content of a hollow Tree or of any other hollow Body whatsoever To perform which first you must find the solid content thereof as though it were not hollow then find the solid content of the Concavity as though it were Massy and subtract it from the whole content and the Remainder will be the solid content of the hollow body PROP. XI How to find the Solid Content of any Solid Body of what Form soever such as Geometry can give no Rule for the measuring thereof To perform which take some convenient Vessel and fill the same to a convenient heighth with clean water and make a Mark just how high the water reacheth then put the Body whose solid content is required into the water which will cause the water to rise above the Mark then take so much of the water out of the Vessel as is raised above the Mark and put the same into an hollow Cube which have in readiness for that purpose then find the solid content of the said Cube so high as the water reacheth which shall be the solid content of the said body required CHAP. III. Of GAVGING GAuging of Vessels is no other than finding the solid content of the concavities in Inches as taught in the last Chapter and Dividing the same by the number of cubick Inches contained in a Gallon of the Liquor contained in the same which according to the Establishment of the Excise for Ale is 282 and for Wine 231 Inches PROP. I. How to Gauge a Cubical Vessel Admit there is a cube-Vessel represented by the Figure D whose side is 16 Inches and it be desired to know how many Gallons of Wine or Ale the same will hold To find which first find the solid content of the said cube in Inches as taught in Proposition the first of the last Chapter which you will find to be 4096 Inches which reduce into Gallons by dividing the same by 282 for Ale or Beer and the Quotient will be 14.5 which are the number of Gallons of Ale

all the VValls of 3 ½ Bricks thick in 1 ½ Brick thick as in the following VVork Then likewise Reduce the Sum of the Third Column 87 Feet 10 Inches after the same Method into 1 Brick and ½ thick which you will find to be 175 Feet 8 Inches as in the following VVork Thus having Reduced the Sum of each Column in this Example into 1 Brick and ½ thick Add them all together and their Sum will be 686 Feet 9 Inches 10 Seconds which is the Content of all the Brick-VVork Required in 1 Brick and ½ thick which Reduce into Rods by Dividing it by 272 the Number of Feet contained in a Rod so will you have 2 Rods and 142 272 of a Rod which is somthing above ½ a Rod for the Content of the Brick-Work to the first floor required as in the following Work What hath been said I suppose is sufficient to Explain the beforegoing Rules so that to give you any more Examples of that kind will be needles But because there is some Walls that have Watet-tables that is Built some 3 some 5 Feet High which Water-table is many times ½ a Brick thick and some times more I shall shew you how to Measure any such Wall To Measure which first find the Content of the Wall as before taught and Reduce the same into Brick and half Thick then find the Content of the Water-Table after the same manner and likewise Reduce it into 1 ½ Brick Thick then Add the Content of the Water-Table to the Content of the Wall and the Sum is the Content of the said Wall with the Water-Table which Reduce into Rods as before Taught Example Admit rhere is a Wall that is 2 Bricks Thick being 91 Feet 6 Inches Long and 17 Feet 3 Inches high which hath a Water-Table ½ a Brick Thick and 3 Feet High whose Content is required To find which first find the Content of the Wall by Multiplying the Heighth into the Length and the Product will be 1578 Feet 4 Inches 6 Seconds which Reduce into 1 ½ Brick Thick by Multiplying it by 4 the Number of half Bricks contained in the given Thickneses and Dividing the Product by 3 the Number of half Bricks contained in 1½ Brick and so will you find the Content of the Wall in Brick and ½ Thick to be 2104 Feet 6 Inches as in the following Work Then find the Content of the Water-Table by Multiplying it's Heigth 3 Feet into the Length of the Wall 91 Feet 6 Inches and the Product will be 274 Feet 6 Inches which Reduce into 1½ Brick Thick which because the Thickness of the said Water-Table is a ½ Brick Thick is done by the Dividing by 3 and the Quotient will be 91 Feet 6 Inches for the Content of the said Water-Table in Brick and ½ as in the Following Work Then Add the Content of the Wall in a Brick and half Thick before found to be 2104 Feet 6 Inches to the Content of the Water-Table in 1½ Brick thick and their Sum will be 2196 Feet for the Content of the Wall Required which you must Reduce into Rods as before taught so will you find 8 Rods and 20 Feet of Brick-Work to be in the said Wall as in the following Work How to Measure Chimnies The Common way allowed by all Measurers is thus girt the Chimney Round below the Mantle-Tree if it be in a Wood Building where the Wall of the House doth go to make the Back of the Chimney and take the girt of the Chimny for the length and the heighth of the Room for the Breadth and then Multiply the one into the other and the Product take for the Content in the same thickness as the Jaumes are on which must be Reduced into 1½ Brick But if the Chimney stands against a Brick Wall then the Wall at the Back being the same with the Wall of the House and therefore being Measured as Part of the Wall they only girt it Round to the Wall and Multiply that into the height of the Room and take the Product for the Content of the Chimney in the same thickness as the Jaums are on which you must Reduce as before into 1 ½ Brick thick Example Admit there is a Chimney whose girt round below the Mantle Tree is 10 Feet 6 Inches and the height of the Room where the same stands 9 Feet the Jaums of which Chimny is 2 Bricks thick whose Content is required To perform which Multiply the girt 10 Feet 6 Inches into the height of the Room 9 Feet and the Product will give 94 Feet 6 Inches for the Content of the said Chimny in the same thickness as the Jaumes is off which ●s 2 Bricks thick which you mu●t Reduce into 1 ½ Brick thick as before directed so will you find the Content thereof in Brick and one half to be 129 Feet as in the following Work Having by some of the before going methods found the Content of all the Chimnies you are next to find the Content of Shafts to perform which you must girt them about in the narrowest place and take that for the breadth and their height for the Length and multiply one into the other and the Product will be the Content the Shaft or Shafts in the same thickness as their breadth is of for you must understand that they are Measured as if they were a Solid Peice of Brick-Work which you must Reduce into 1 ½ Brick thick How to Measure Tyling Tyling is Measured by the Square Containing 100 Feet and is in every respect the same with that of Roofing taught in Measuring Carpenters Work Example Admit there is a Roof of a Building that is Tyled the length of which Building from the out-side of the Wall at one End to the out side of the Wall at the other End is 15 Feet and the Breadth from the out-side of the one Wall to the out-side of the other 9 Feet 6 Inches and it be required to find the Content of the said Tyling which you will find to be 2 Squares 1 Foot 3 Inches and 81 100 of an Inch. Thus much of Measuring of Artificers Work relating to Building I presume what hath been said is sufficient for the meanest Capacity to understand how to Measure any sort of Building Timber Stone or any sort of Body whatsoever FINIS Advertisement of Choice new Books lately Printed for and Published by Tho. Salusbury at the King's Armes next St. Dunstan's Church in Fleetstreet THE Gazetteer's or Newsman's Interpreter Being a Geographical Index of all the Considerable Cities Patriarchships Bishopricks Vniversities Dukedoms Earldoms and such like Imperial and Hance Towns Ports Forts Castles c. in Europe Shewing in what Kingdoms Provinces and Counties they are in to what Prince they are now Subject upon or nigh what Rivers Bays Seas Mountains c. they stand their Distances in English Miles from several other Places of Note with their Longitude and Latitude according to the best and approved Maps Of special use for the true Understanding of all Modern Histories of Vrope as well as the Present Affares and for the Conveniency of Cheapness and Pocket-Carriage Explained by Abbreviations and Figures By Law Eachard A. B. of Christ's Colledge in Cambridge 2. The History of the Late Great Revolution in England and Scotland With the Causes and Means by which it was Accomplished Together with a Particular Account of the Extraordinary Occurrencies which happned thereupon As likewise the Settlement of both the Kingdoms under Their Most Serene Majesties King William and Queen Mary Price 5 Shillings 3. Zingis or Love prefer'd before Duty a Novel Translated out of French Price bound 1 Shilling 4. Arithmetical Rules digested and Contracted made plain and easie for the help and benefit of the Memory Very necessary for all Gentlemen and Tradesmen as for Youth and Apprentices in Merchantile Affairs By Arthur Leadbetter School-Master in Bury St. Edmonds in Suffolk Twelves Price 1. s. 5. Chimichus Rationalis or the Fundamental Grounds of the Chimical Art rationally stated and demonstrated by various Examples in Destillation Rectification and Exaltation of Wines Spirits Tinctures Oyles Salts Powders and Oleosums in such a Method as to retain the Specifick Vertues of Concerts in the greatest power and Force By W. Yworth Spagyrick Physitian and Philosopher by Fire Price bound 1 s. 6. A new Art of Brewing Beer Ale and other sorts of Liquors so as to render them more healthful to the Body and agreeable to Nature and to keep the longer from sowring with less Trouble and Charge than generally practised To which is added the Art of making Mault The third Edition By Tho. Tryon Student in Physick Twelves Price bound 1 s. 7. A new Art of making above 20 sorts of Wines Brandy and other Spirits more pleasant and agreeable to the English Constitution than those of France compliant to the late Act of Parlament and Illustrated by the Doctrine of Fermentation and Distillation by the various Examples on the Growth and Products of this Island By W. Y. Medicinal Professor 12 s. Price bound 1 s. 6 d. 8. Wisdoms Dictares or Rules Physical and Moral for preserving the Health of the Body and the Peace of the Mind Fit to be regarded and practised by all that would enjoy the Blessing of this World To which is added a Bill of Fare of 75 noble Dishes of Excellent Food without either Fish or Flesh 〈◊〉 Th● Tryon Student in 〈…〉